posted by Kim on .
A mixture of ethyne gas (C2H2) and methane gas (CH4) occupied a certain volume at a total pressure of 16.8 kPa. When the sample burned, the products were CO2 gas and H2O vapor. The CO2 was collected and its pressure found to be 25.2 kPa in the same volume and same temperature as the original mixture. What percentage of the original mixture was methane?
Check my thinking on this.
The gas starts as 16.8 kPa Ptotal.
It ends up as 25.2 kPa Ptotal.
Look at the equations.
C2H2 + 5/2 O2 = 2CO2 + H2O
CH4 + 2O2 ==> CO2 + 2H2O
You can see that 1 mol C2H2 produces 2 moles CO2 while 1 mol CH4 produces 1 mol CO2. IF WE HAD 100% CH4, we would expect to obtain 16.8 kPa CO2. The difference (25.2-16.8 = 8.4) must be due to the C2H2 present. Then 16.8 = Ptotal and 8.4 = PC2H2 and 16.8-8.4 = 8.4 PCH4. So we are starting with 8.4 kPa CH4 and 8.4 kPa C2H2 or 50% each.
wrong, answer is that the methane makes up 46% of the gases being burned. (I have the chem textbook that the Q comes from).